An electron tube was sealed off during manufacture at a pressure of `1.2xx10^(-7)` mm of mercury at `27^(@)C.` Its volume is 100 `cm^(3)`. The number of molecules that remain in the tube is

Options

A. `2xx10^(16)`
B. `3xx10^(15)`
C. `3.86xx10^(11)`
D. `5xx10^(11)`

Explanation

To find the number of molecules in the electron tube, we can use the ideal gas equation, which is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

However, since we want to find the number of molecules, we can express the number of moles \( n \) in terms of the number of molecules \( N \) using Avogadro's number \( N_A \):

\[ n = \frac{N}{N_A} \]

Substituting this into the ideal gas equation gives:

\[ PV = \frac{N}{N_A} RT \]

Rearranging for \( N \):

\[ N = \frac{P V N_A}{RT} \]

### Step 1: Convert the given values
1. **Pressure**: The pressure is given as \( 1.2 \times 10^{-7} \) mm of mercury. We need to convert this to Pascals (Pa) using the conversion factor \( 1 \text{ mmHg} = 133.322 \text{ Pa} \).
\[ P = 1.2 \times 10^{-7} \text{ mmHg} \times 133.322 \text{ Pa/mmHg} = 1.598 \times 10^{-5} \text{ Pa} \]

2. **Volume**: The volume is given as \( 100 \text{ cm}^3 \). Convert this to cubic meters (m³):
\[ V = 100 \text{ cm}^3 = 100 \times 10^{-6} \text{ m}^3 = 1 \times 10^{-4} \text{ m}^3 \]

3. **Temperature**: The temperature is given as \( 27^{\circ}C \). Convert this to Kelvin:
\[ T = 27 + 273.15 = 300.15 \text{ K} \approx 300 \text{ K} \]

### Step 2: Use the ideal gas equation to find the number of molecules
Now we can substitute the values into the rearranged ideal gas equation:

- \( R \) (universal gas constant) = \( 8.314 \text{ J/(mol K)} \)
- \( N_A \) (Avogadro's number) = \( 6.022 \times 10^{23} \text{ molecules/mol} \)

Substituting into the equation:

\[ N = \frac{(1.598 \times 10^{-5} \text{ Pa}) \times (1 \times 10^{-4} \text{ m}^3) \times (6.022 \times 10^{23} \text{ molecules/mol})}{(8.314 \text{ J/(mol K)}) \times (300 \text{ K})} \]

### Step 3: Calculate the number of molecules
Calculating the numerator:

\[ N = \frac{(1.598 \times 10^{-5}) \times (1 \times 10^{-4}) \times (6.022 \times 10^{23})}{(8.314) \times (300)} \]

Calculating the denominator:

\[ 8.314 \times 300 = 2494.2 \]

Now, calculating the entire expression:

\[ N = \frac{(1.598 \times 10^{-9}) \times (6.022 \times 10^{23})}{2494.2} \]

Calculating the numerator:

\[ 1.598 \times 10^{-9} \times 6.022 \times 10^{23} \approx 9.619 \times 10^{14} \]

Now dividing by the denominator:

\[ N \approx \frac{9.619 \times 10^{14}}{2494.2} \approx 3.86 \times 10^{11} \]

Thus, the number of molecules in the tube is approximately:

\[ \boxed{3.86 \times 10^{11}} \]